A) \[Ba{{({{N}_{3}})}_{2}}(s)\to Ba(C)+3{{N}_{2}}(g)\]
B) \[{{(N{{H}_{4}})}_{2}}C{{r}_{2}}{{O}_{7}}(s)\to {{N}_{2}}(g)+4{{H}_{2}}O(g)+C{{r}_{2}}{{O}_{3}}(s)\]
C) \[2N{{H}_{3}}(g)\to {{N}_{2}}(g)+3{{H}_{2}}(g)\]
D) \[2N{{H}_{4}}N{{O}_{3}}(s)\to 2{{N}_{2}}(g)+4{{H}_{2}}O(g)+{{O}_{2}}(g)\]
Correct Answer: C
Solution :
For per gram of reactant, the maximum quantity \[{{N}_{2}}\]of gas is produced in the thermal decomposition\[2N{{H}_{3}}(g)\to {{N}_{2}}(g)+3{{H}_{2}}(g)\] (A) Molar mass of \[Ba{{({{N}_{3}})}_{2}}(s)=221g/mol.1mole\]of \[Ba{{({{N}_{3}})}_{2}}(s)\] will give 3 moles of \[{{N}_{2}}\] \[\frac{1g}{221g/mol}\]moles of \[Ba{{({{N}_{3}})}_{2}}(s)\]will give \[3\times \frac{1}{221}=0.014\]moles of \[{{N}_{2}}\] (B) Molar mass of \[{{(N{{H}_{4}})}_{2}}C{{r}_{2}}{{O}_{7}}=252g/mol.\]\[1mole\] of \[{{(N{{H}_{4}})}_{2}}C{{r}_{2}}{{O}_{7}}\] will give 1 mole of \[{{N}_{2}}\] \[\frac{1g}{252g/mol}\]moles of \[{{(N{{H}_{4}})}_{2}}C{{r}_{2}}{{O}_{7}}\] will give \[1\times \frac{1}{252}=0.039\]moles of\[{{N}_{2}}\] (C) Molar mass of \[N{{H}_{3}}=17g/mol,2mole\] of \[N{{H}_{3}}\] will give 1 mole of\[{{N}_{2}}\] \[\frac{1g}{17g/mol}\] moles of \[N{{H}_{3}}\] will give \[\frac{1}{2\times 17}=0.0297\] moles of \[{{N}_{2}}\] (D) Molar mass of \[N{{H}_{4}}N{{O}_{3}}=80g/mol.1mole\] of \[N{{H}_{4}}N{{O}_{3}}\] will give 1 mole of \[{{N}_{2}}\] \[\frac{1g}{80g/mol}\]moles of \[N{{H}_{4}}N{{O}_{3}}\] will give \[1\times \frac{1}{80}=0.0125\] moles of \[{{N}_{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
